Is there a norm for $\nabla$ operator? What is the meaning of $\sqrt{\Delta}$?

Question

Considering the del operator in cartesian coordinate. $$\nabla=\left(\dfrac{\partial}{\partial x_1},...,\dfrac{\partial}{\partial x_n}\right)$$ Since technically it's a vector, does its norm have any meaning?
I considered is "norm" as the square root of the Laplacian:
$$\left\Vert\nabla\right\Vert_{\mathbb{R}^n}=\sqrt{\Delta}=\sqrt{\dfrac{\partial^2}{\partial x_1^2}+...+\dfrac{\partial^2}{\partial x_n^2}}$$
Is it correct to think of this operator as "the operator that applied twice behaves like the Laplacian"?
If this is wrong, it is possible to make the norm of a vector of operators?
If yes, what are the properties of this norm and what normed space are we in?
In this case I considered $\mathbb{R}^n$ and technically this is not a norm since the result is not a non-negative number, so I was curious to know:

1. If anyway this operator makes sense (in case what is the meaning).
2. If there is a space that can associate a numeric value to the norm of this operator (even if the norm was infinite causing the operator to be unbounded).
3. If this space exists, does the result depend on the coordinate system i'm using and the length of the vector?

Answer 1

Yes, $\sqrt{\Delta}$ is supposed to be an operator which, when applied twice, is $\Delta$. This is not a norm; the reason the formulas have the same form is because both are square roots of rotation-invariant quadratics. I believe this kind of operator is called a Dirac operator, and Dirac tried using it to describe a relativistic quantum theory. To do it, he needed to extend scalars from complex numbers to Clifford algebras (the associative algebra generated by multiple anticommuting square roots of $\pm1$; sign conventions vary).

That said, you can talk about norms of operators, generalizing matrix norms. For example, the Frobenius norm is $\|A\|^2=\mathrm{tr}(A^\dagger A)=\sum_{ij}|a_{ij}|^2$, which generalizes to infinite dimensions and is then called the Hilbert-Schmidt norm (or inner product: $\langle A,B\rangle=\mathrm{tr}(A^\dagger B)$, if we use the physicists' convention of being left-conjugate-linear). For Hilbert spaces, we're talking about "trace class" functions in order to take the trace, which may be computed as $\mathrm{tr}(S)=\sum\langle Se_i,e_i\rangle=\sum_{ij}|\langle Ae_i,e_j\rangle|^2$ for $S=A^\dagger A$ and a choice of orthonormal (Hilbert) basis $\{e_i\}$. The trace, and thus the norm, are coordinate-independent, but are of course necessarily tied to the inner product.

Using regularization, it is also possible to extend this to other operators, including differential, but I'm not sure how sensible this is. In the matrix case, we know $\|A\|^2=\sum |\lambda|^2$, where the eigenvalues $\lambda$ are counted with multiplicity, and we can generalize this to operators like $\Delta$. This will depend on the manifold in question, of course. For the circle $S^1$, the eigenfunctions of $\partial^2/\partial\theta^2$ are $f(z)=z^n$ with eigenvalues $\lambda=n^2$, and we can write $\sum_{n\in\Bbb Z}n^4=0+2\zeta(-4)=0$, which seems weird. What is done in practice is regularizing the determinant of an operator, i.e. $\det S=\prod\lambda=\exp(\sum\ln\lambda)$ which gives $\det(\partial^2/\partial\theta^2)=e^{-2\zeta'(0)}=2\pi$.